Съдържание

Условие Отговори Решение

OPhO 2026, Единствен кръг — Задача 10. Countless Collisions

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от OPhO 2026

2026 Online Physics Olympiad · 21 август 2026 г.

Last updated

Last updated: 22nd August 03:35 UTC

Sponsors

This competition would not be possible without the help of our sponsors.

Corrections (all times are in UTC)

Q8: Cleared up wording (21/08 04:40 and 04:52)
Q3: improved wording (21/08 08:49)
Q19: Improved wording (21/08 09:42)
Q16: Attempts reset (21/08 22:00), Error bound increased (22/8 02:30)
Q1: Improved wording (22/08 03:35)

Указания

Format. The OPhO is a team-based competition (3 members at most per team). Unlike previous years, the 2026 OPhO consists of a single round held from August 21, 12 am (UTC), to August 23, 11:59 pm (UTC). The contest is made up of three parts, all running concurrently over the full contest window:

Short Answer — problems requiring numerical answers, submitted on the OPhO portal;
Long Answer — problems requiring full written solutions, submitted on the OPhO portal;
Simulation — computational problems requiring teams to run a provided executable locally, with solutions submitted on the OPhO portal.

All teams have access to all three parts for the entire duration of the contest. Submissions are made on the OPhO portal https://opho.physoly.org/ using the team login, and may be made at any time within the contest window.

Short Answer. Teams submit numerical answers on the portal, with up to three attempts per problem. Every answer submission must be accompanied by a file upload, on the portal, of the working used to obtain that answer. This working will not be graded, but submissions without it will not be accepted. The base score for each problem depends on the number of teams that successfully answer it, each incorrect attempt multiplies the base score by a factor of 0.8, and there is no time factor. The score for an individual problem follows:

w(N) = (0.8)^i\left[e^{-\frac{x(N)}{200}} + e^{-\frac{x(N)^2}{100}} + 1\right],

where $i \in \{0, 1, 2\}$ represents whether a team got it on the first, second, or third try, $N$ represents the number of teams that answer the problem correctly, and $x(N)$ is the sum of $(0.8)^i$ across all teams that have solved the problem. The base score is independent of problem number, but expect questions to be ordered roughly by difficulty. A team’s raw Short Answer total is the sum of these values, and all raw totals are then scaled linearly such that the top team receives 55 points.

Long Answer. Teams submit full solutions on the portal. Rubrics are predetermined, and solutions will be reviewed to determine the proper number of points to give. The points breakdown is given per problem, for a total of 35 points. It is recommended that participants write their solutions in LaTeX. Handwritten solutions (or a combination of both) are accepted too; if participants have more than one photo of a handwritten solution (jpg, png, etc.), it is required that they be organized in the correct order in a pdf before submitting.

Simulation. Teams will need to be able to run a provided executable locally. Solutions are submitted on the portal and will be graded against predetermined rubrics. The points breakdown is given per problem, for a total of 10 points.

Scoring. A team’s cumulative score is the sum of its scaled Short Answer score (out of 55 points), Long Answer score (out of 35 points), and Simulation score (out of 10 points), for a maximum score of 100 points.

Permitted Resources. Use of the internet is allowed, including textbooks, published papers, calculators, and computer algebra systems. However, the use of AI tools of any kind is strictly prohibited. This includes, but is not limited to, large language models (e.g. ChatGPT, Claude, Gemini), AI coding assistants, AI-generated search summaries, and any other AI-powered assistance.

Dishonest Behavior. Please refrain from cheating in any form. This includes, but is not limited to, creating alternate accounts to gain additional attempts, sharing answers with other teams, posting problems online, utilizing AI tools as described above, or employing any other method to gain an unfair advantage.

We have previously caught numerous teams engaging in such practices, and pursuing dishonest tactics provides no genuine sense of security. Any teams found to be acting maliciously will be notified via email, along with the opportunity to appeal. Failing to provide a valid appeal will result in a ban from participating in any future Physoly competitions. Our utmost priority is to ensure an enjoyable contest experience for all participants, and we remain committed to preventing any disruption to the fun for others.

Awards and Results. The top competitors will be recognized on the website. No cash prizes will be awarded this year.

List of Constants

Proton mass, $m_p = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Neutron mass, $m_n = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Electron mass, $m_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}$
Avogadro’s constant, $N_0 = 6.02 \cdot 10^{23}\ \mathrm{mol^{-1}}$
Universal gas constant, $R = 8.31\ \mathrm{J/(mol \cdot K)}$
Boltzmann’s constant, $k_B = 1.38 \cdot 10^{-23}\ \mathrm{J/K}$
Electron charge magnitude, $e = 1.60 \cdot 10^{-19}\ \mathrm{C}$
1 electron volt, $1\ \mathrm{eV} = 1.60 \cdot 10^{-19}\ \mathrm{J}$
Speed of light, $c = 3.00 \cdot 10^{8}\ \mathrm{m/s}$
Universal Gravitational constant, $G = 6.67 \cdot 10^{-11}\ \mathrm{(N \cdot m^2)/kg^2}$
Solar Mass $M_\odot = 1.988 \cdot 10^{30}\ \mathrm{kg}$
Acceleration due to gravity, $g = 9.8\ \mathrm{m/s^2}$
1 unified atomic mass unit, $1\ \mathrm{u} = 1.66 \cdot 10^{-27}\ \mathrm{kg} = 931\ \mathrm{MeV/c^2}$
Planck’s constant, $h = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s} = 4.41 \cdot 10^{-15}\ \mathrm{eV \cdot s}$
Permittivity of free space, $\epsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}$
Coulomb’s law constant, $k = \frac{1}{4\pi\epsilon_0} = 8.99 \cdot 10^{9}\ \mathrm{(N \cdot m^2)/C^2}$
Permeability of free space, $\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{T \cdot m/A}$
Magnetic constant, $\frac{\mu_0}{4\pi} = 1 \cdot 10^{-7}\ \mathrm{(T \cdot m)/A}$
1 atmospheric pressure, $1\ \mathrm{atm} = 1.01 \cdot 10^{5}\ \mathrm{N/m^2} = 1.01 \cdot 10^{5}\ \mathrm{Pa}$
Wien’s displacement constant, $b = 2.9 \cdot 10^{-3}\ \mathrm{m \cdot K}$
Stefan-Boltzmann constant, $\sigma = 5.67 \cdot 10^{-8}\ \mathrm{W/m^2/K^4}$

Условие

This problem is one-dimensional, with all objects constrained to the $x$ -axis. Consider a block of mass $M$ with initial velocity $-v_0$ and position $x = L$ at $t = 0$ . A dispenser is positioned at $x = 0$ , given an initial velocity $v_1$ , and uniformly releases small balls of mass $m$ and initial velocity $u$ relative to the lab frame in time intervals of $k$ starting at $t = 0$ . If all collisions are elastic, find the minimum distance between the dispenser and block after a very long time has elapsed. Neglect all collisions with the dispenser and assume that the dispenser always moves at the same velocity. Report the answer in meters.

Use the following parameters: $M = 59290\ \mathrm{kg}$ , $v_0 = 5.929\ \mathrm{m/s}$ , $v_1 = 2.34\ \mathrm{m/s}$ , $L = 5929\ \mathrm{m}$ , $m = 0.5929\ \mathrm{kg}$ , $k = 0.005929\ \mathrm{s}$ , $u = 5.929\ \mathrm{m/s}$ .

Отговори

Решение

Work in the big block’s frame and let its velocity relative to the lab frame be $v$ . In this frame, each small block comes in with velocity $u - v$ and has velocity $v - u$ after the collision. Boosting back to the lab frame after the collision gives velocity $2v - u$ for the small block. Now we conserve energy in the lab frame, letting $v_f$ represent the velocity of block $M$ after the collision.

\frac{1}{2}mu^2 + \frac{1}{2}Mv^2 = \frac{1}{2}Mv_f^2 + \frac{1}{2}m(2v - u)^2

Mv_f^2 = Mv^2 + 4muv - 4mv^2

v_f = \sqrt{v^2 + 4muv/M - 4mv^2/M} \approx v(1 + 2mu/(Mv) - 2m/M)

Since all collisions are elastic and the balls are identical, we can treat the balls as freely passing through each other. Let $v_M$ be the velocity of $M$ . Due to the doppler shift, the frequency $f$ at which balls collide with $M$ is

f = \frac{1}{k}(\frac{u - v_M}{u - v_1})

Use the previously derived formula and let the function $m(t)$ represent the total mass which has collided with the big block as a function of time $t'$ , where the first collision happens at $t' = 0$ :

\frac{\mathrm{d}v_M}{\mathrm{d}m(t')} = 2u/M - 2v_M/M

\int_{-v_0}^{v_M}\frac{1}{u - v_M}\mathrm{d}v_M = \int_0^{m(t)}\frac{2}{M}\mathrm{d}m(t')

\ln\left(\frac{u + v_0}{u - v_M}\right) = 2m(t')/M

v_M = u - (u + v_0)e^{-\frac{2m(t')}{M}}

We know that

\frac{\mathrm{d}m(t')}{\mathrm{d}t'} = mf = \frac{m}{k}(\frac{u - v_M}{u - v_1})

\int_0^{m(t)}\frac{\mathrm{d}m(t)}{(u + v_0)e^{-\frac{2m(t')}{M}}} = \int_0^{t'}\frac{m\mathrm{d}t'}{k(u - v_1)}

\frac{M}{2}(e^{\frac{2m(t')}{M}} - 1) = \frac{m(u + v_0)t'}{k(u - v_1)}

e^{\frac{2m(t')}{M}} = 1 + \frac{2m(u + v_0)t'}{Mk(u - v_1)}

Thus,

v_M = u - \frac{u + v_0}{1 + \frac{2m(u + v_0)t'}{Mk(u - v_1)}}

The distance $d$ between the dispenser and the block as a function of $t'$ is

d = L - v_1(\frac{L}{u + v_0}) - v_0(\frac{L}{u + v_0}) - v_1 t' + \int_0^{t'} v_M\mathrm{d}t'

d = L - \frac{(v_0 + v_1)L}{u + v_0} - v_1 t' + ut' - \frac{Mk(u - v_1)}{2m}\ln\left(1 + \frac{2m(u + v_0)t'}{Mk(u - v_1)}\right)

Now, there are a number of ways to proceed: we can visually identify the minimum $d$ on a graph, set the derivative of $d$ with respect to $t'$ equal to zero, solve for $t'$ , and plug back into the expression for $d$ , or set the previous expression for $v_M$ equal to $v_1$ before solving for $t'$ and proceeding as the previous method. Regardless of the method chosen, the minimum should be found to occur at $t' \approx 207$ s and $d \approx 1265$ m.

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf